To study undamped oscillations under the influence of the external force $F(t)=F_0\cos \omega t\text{,}$ we set $c=0$ in Eq. (1), and thereby begin with the equation

whose complementary function is $x_c=c_1\cos {\omega }_{0}t+c_2\sin {\omega }_{0}t\text{.}$ Here

(as in Eq. (9) of Section 5.4) is the (circular) natural frequency of the mass–spring system. The fact that the angle ${\omega }_{0}t$ is measured in (dimensionless) radians reminds us that if t is measured in seconds (s), then ${\omega }_0$ is measured in radians per second—that is, in inverse seconds (${\text{s}}^{-1}$). Also recall from Eq. (14) in Section 5.4 that division of a circular frequency $\omega$ by the number $2\pi$ of radians in a cycle gives the corresponding (ordinary) frequency $\nu =\omega /2\pi$ in Hz (hertz $=$ cycles per second).

Let us assume initially that the external and natural frequencies are unequal: $\omega \ne {\omega }_0\text{.}$ We substitute $x_p=A\cos \omega t$ in Eq. (4) to find a particular solution. (No sine term is needed in $x_p$ because there is no term involving $x\text{′}$ on the left-hand side in Eq. (4).) This gives

so

and thus

Therefore, the general solution $x=x_c+x_p$ is given by

where the constants $c_1$ and $c_2$ are determined by the initial values $x(0)$ and $x\text{′}(0)\text{.}$ Equivalently, as in Eq. (12) of Section 5.4, we can rewrite Eq. (7) as

so we see that the resulting motion is a superposition of two oscillations, one with natural circular frequency ${\omega }_0\text{,}$ the other with the frequency $\omega$ of the external force.

Example 1

Suppose that $m=1,k=9,F_0=80\text{,}$ and $\omega =5\text{,}$ so the differential equation in (4) is

$$x″+9x=80\cos 5t.$$

Find x(t) if $x(0)=x\text{′}(0)=0$.

Solution

Here the natural frequency ${\omega }_0=3$ and the frequency $\omega =5$ of the external force are unequal, as in the preceding discussion. First we substitute $x_p=A\cos 5t$ in the differential equation and find that $-25A+9A=80\text{,}$ so that $A=-5\text{.}$ Thus a particular solution is

$$x_p(t)=-5\cos 5t.$$

The complementary function is $x_c=c_1\cos 3t+c_2\sin 3t\text{,}$ so the general solution of the given nonhomogeneous equation is

$$x(t)=c_1\cos 3t+c_2\sin 3t-5\cos 5t,$$

with derivative

$$x\text{′}(t)=-3c_1\sin 3t+3c_2\cos 3t+25\sin 5t.$$

The initial conditions $x(0)=0$ and $x\text{′}(0)=0$ now yield $c_1=5$ and $c_2=0\text{,}$ so the desired particular solution is

$$x(t)=5\cos 3t-5\cos 5t.$$

As indicated in Fig. 5.6.2, the period of x(t) is the least common multiple $2\pi$ of the periods $2\pi /3$ and $2\pi /5$ of the two cosine terms.

If we impose the initial conditions $x(0)=x\text{′}(0)=0$ on the solution in (7), we find that

so the particular solution is

The trigonometric identity $2\sin A\sin B=\cos (A-B)-\cos (A+B)\text{,}$ applied with $A=\frac{1}{2}({\omega }_0+\omega )t$ and $B=\frac{1}{2}({\omega }_0-\omega )t\text{,}$ enables us to rewrite Eq. (9) in the form

Suppose now that $\omega \approx {\omega }_0\text{,}$ so that ${\omega }_0+\omega$ is very large in comparison with $|{\omega }_0-\omega |\text{.}$ Then $\sin \frac{1}{2}({\omega }_0+\omega )t$ is a rapidly varying function, whereas $\sin \frac{1}{2}({\omega }_0-\omega )t$ is a slowly varying function. We may therefore interpret Eq. (10) as a rapid oscillation with circular frequency $\frac{1}{2}({\omega }_0+\omega )\text{,}$

but with a slowly varying amplitude

Example 2

With $m=0.1,F_0=50,{\omega }_0=55\text{,}$ and $\omega =45\text{,}$ Eq. (10) gives

$$x(t)=\sin 5t\sin 50t\text{.}$$

Figure 5.6.3 shows the corresponding oscillation of frequency $\frac{1}{2}({\omega }_0+\omega )=50$ that is “modulated” by the amplitude function $A(t)=\sin 5t$ of frequency $\frac{1}{2}({\omega }_0-\omega )=5$.

A rapid oscillation with a (comparatively) slowly varying periodic amplitude exhibits the phenomenon of beats. For example, if two horns not exactly attuned to one another simultaneously play their middle C, one at ${\omega }_0/(2\pi )=258$ Hz and the other at $\omega /(2\pi )=254$ Hz, then one hears a beat—an audible variation in the amplitude of the combined sound—with a frequency of

Looking at Eq. (6), we see that the amplitude A of $x_p$ is large when the natural and external frequencies ${\omega }_0$ and $\omega$ are approximately equal. It is sometimes useful to rewrite Eq. (5) in the form

where $F_0/k$ is the static displacement of a spring with constant k due to a constant force $F_0\text{,}$ and the amplification factor $\rho$ is defined to be

It is clear that $\rho \to +\infty$ as $\omega \to {\omega }_0\text{.}$ This is the phenomenon of resonance—the increase without bound (as $\omega \to {\omega }_0$) in the amplitude of oscillations of an undamped system with natural frequency ${\omega }_0$ in response to an external force with frequency $\omega \approx {\omega }_0$.

We have been assuming that $\omega \ne {\omega }_0\text{.}$ What sort of catastrophe should one expect if $\omega$ and ${\omega }_0$ are precisely equal? Then Eq. (4), upon division of each term by m, becomes

Because $\cos {\omega }_{0}t$ is a term of the complementary function, the method of undetermined coefficients calls for us to try

We substitute this in Eq. (13) and thereby find that $A=0$ and $B=F_0/(2m{\omega }_0)\text{.}$ Hence the particular solution is

The graph of $x_p(t)$ in Fig. 5.6.4 (in which $m=1,F_0=100\text{,}$ and ${\omega }_0=50$) shows vividly how the amplitude of the oscillation theoretically would increase without bound in this case of pure resonance, $\omega ={\omega }_0\text{.}$ We may interpret this phenomenon as reinforcement of the natural vibrations of the system by externally impressed vibrations at the same frequency.

In practice, a mechanical system with very little damping can be destroyed by resonance vibrations. A spectacular example can occur when a column of soldiers marches in step over a bridge. Any complicated structure such as a bridge has many natural frequencies of vibration. If the frequency of the soldiers’ cadence is approximately equal to one of the natural frequencies of the structure, then—just as in our simple example of a mass on a spring—resonance will occur. Indeed, the resulting resonance vibrations can be of such large amplitude that the bridge will collapse. This has actually happened—for example, the collapse of Broughton Bridge near Manchester, England, in 1831—and it is the reason for the now-standard practice of breaking cadence when crossing a bridge. Resonance may have been involved in the 1981 Kansas City disaster in which a hotel balcony (called a skywalk) collapsed with dancers on it. The collapse of a building in an earthquake is sometimes due to resonance vibrations caused by the ground oscillating at one of the natural frequencies of the structure; this happened to many buildings in the Mexico City earthquake of September 19, 1985. On occasion an airplane has crashed because of resonant wing oscillations caused by vibrations of the engines. It is reported that for some of the first commercial jet aircraft, the natural frequency of the vertical vibrations of the airplane during turbulence was almost exactly that of the mass–spring system consisting of the pilot’s head (mass) and spine (spring). Resonance occurred, causing pilots to have difficulty in reading the instruments. Large modern commercial jets have different natural frequencies, so that this resonance problem no longer occurs.

The avoidance of destructive resonance vibrations is an ever-present consideration in the design of mechanical structures and systems of all types. Often the most important step in determining the natural frequency of vibration of a system is the formulation of its differential equation. In addition to Newton’s law $F=ma\text{,}$ the principle of conservation of energy is sometimes useful for this purpose (as in the derivation of the pendulum equation in Section 5.4). The following kinetic and potential energy formulas are often useful.

1. Kinetic energy: $T=\frac{1}{2}m{v}^2$ for translation of a mass m with velocity v;

2. Kinetic energy: $T=\frac{1}{2}I{\omega }^2$ for rotation of a body of a moment of inertia I with angular velocity $\omega$;

3. Potential energy: $V=\frac{1}{2}k{x}^2$ for a spring with constant k stretched or compressed a distance x;

4. Potential energy: $V=mgh$ for the gravitational potential energy of a mass m at height h above the reference level (the level at which $V=0$), provided that g may be regarded as essentially constant.

Example 4

Rolling disk Find the natural frequency of a mass m on a spring with constant k if, instead of sliding without friction, it is a uniform disk of radius a that rolls without slipping, as shown in Fig 5.6.5.

Solution

With the preceding notation, the principle of conservation of energy gives

$$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}k{x}^2=E$$

where E is a constant (the total mechanical energy of the system). We note that $v=a\omega$ and recall that $I=m{a}^2/2$ for a uniform circular disk. Then we may simplify the last equation to

$$\frac{3}{4}m{v}^2+\frac{1}{2}k{x}^2=E\text{.}$$

Because the right-hand side of this equation is constant, differentiation with respect to t (with $v=x\text{′}$ and $v\text{′}=x″$) now gives

$$\frac{3}{4}mx\text{′}x″+kxx\text{′}=0.$$

We divide each term by $\frac{3}{2}mx\text{′}$ to obtain

$$x^{″}+{\displaystyle \frac{2k}{3m}}x=0.$$

Thus the natural frequency of horizontal back-and-forth oscillation of our rolling disk is $\sqrt{2k/3m}\text{,}$ which is $\sqrt{\mathrm{2/3}}\approx 0.8165$ times the familiar natural frequency $\sqrt{k/m}$ of a mass on a spring that is sliding without friction rather than rolling without sliding. It is interesting (and perhaps surprising) that this natural frequency does not depend on the radius of the disk. It could be either a dime or a large disk with a radius of one meter (but of the same mass).

Example 5

Car suspension Suppose that a car oscillates vertically as if it were a mass $m=800\text{ kg}$ on a single spring (with constant $k=7\times {10}^4$ N/m), attached to a single dashpot (with constant $c=3000\text{ N}·\text{s/m}$). Suppose that this car with the dashpot disconnected is driven along a washboard road surface with an amplitude of 5 cm and a wavelength of $L=10$ m (Fig. 5.6.6). At what car speed will resonance vibrations occur?

Solution

We think of the car as a unicycle, as pictured in Fig. 5.6.7. Let x(t) denote the upward displacement of the mass m from its equilibrium position; we ignore the force of gravity, because it merely displaces the equilibrium position as in Problem 9 of Section 5.4. We write the equation of the road surface as

$$y=a\cos {\displaystyle \frac{2\pi s}{L}}(a=0.05\text{ m, }L=10\text{ m}).$$ (15)

When the car is in motion, the spring is stretched by the amount $x-y\text{,}$ so Newton’s second law, $F=ma\text{,}$ gives

$$mx″=-k(x-y);$$

that is,

$$mx″=+kx=ky.$$ (16)

If the velocity of the car is v, then $s=vt$ in Eq. (15), so Eq. (16) takes the form

$$mx″=+kx=ka\cos {\displaystyle \frac{2\pi vt}{L}}.$$ (16′)

This is the differential equation that governs the vertical oscillations of the car. In comparing it with Eq. (4), we see that we have forced oscillations with circular frequency $\omega =2\pi v/L\text{.}$ Resonance will occur when $\omega ={\omega }_0=\sqrt{k/m}\text{.}$ We use our numerical data to find the speed of the car at resonance:

$$v=\frac{L}{2\pi }\sqrt{\frac{k}{m}}=\frac{10}{2\pi }\sqrt{{\displaystyle \frac{7\times {10}^4}{800}}}\approx 14.89(\text{m}/\text{s});$$

that is, about 33.3 mi/h (using the conversion factor of 2.237 mi/h per m/s).

In real physical systems there is always some damping, from frictional effects if nothing else. The complementary function $x_c$ of the equation

is given by Eq. (19), (20), or (21) of Section 5.4, depending on whether $c>c_{\mathrm{cr}}=\sqrt{4km},c=c_{\mathrm{cr}}\text{,}$ or $c<c_{\mathrm{cr}}\text{.}$ The specific form is not important here. What is important is that, in any case, these formulas show that $x_c(t)\to 0$ as $t\to +\infty \text{.}$ Thus $x_c$ is a transient solution of Eq. (17)—one that dies out with the passage of time, leaving only the particular solution $x_p$.

The method of undetermined coefficients indicates that we should substitute

in Eq. (17). When we do this, collect terms, and equate coefficients of $\cos \omega t$ and $\sin \omega t\text{,}$ we obtain the two equations

that we solve without difficulty for

If we write

as usual, we see that the resulting steady periodic oscillation

has amplitude

Now (19) implies that $\sin \alpha =B/C>0\text{,}$ so it follows that the phase angle $\alpha$ lies in the first or second quadrant. Thus

so

(whereas $\alpha =\pi /2$ if $k=m{\omega }^2$).

Note that if $c>0\text{,}$ then the “forced amplitude”—defined as a function $C(\omega )$ by (21)—always remains finite, in contrast with the case of resonance in the undamped case when the forcing frequency $\omega$ equals the critical frequency ${\omega }_0=\sqrt{k/m}\text{.}$ But the forced amplitude may attain a maximum for some value of $\omega \text{,}$ in which case we speak of practical resonance. To see if and when practical resonance occurs, we need only graph C as a function of $\omega$ and look for a global maximum. It can be shown (Problem 27) that C is a steadily decreasing function of $\omega$ if $c\geqq \sqrt{2km}\text{.}$ But if $c<\sqrt{2km}\text{,}$ then the amplitude of C attains a maximum value—and so practical resonance occurs—at some value of $\omega$ less than ${\omega }_0\text{,}$ and then approaches zero as $\omega \to +\infty \text{.}$ It follows that an underdamped system typically will undergo forced oscillations whose amplitude is

• Large if $\omega$ is close to the critical resonance frequency;

• Close to $F_0/k$ if $\omega$ is very small;

• Very small if $\omega$ is very large.

Here we investigate forced vibrations of the mass–spring–dashpot system with equation

To simplify the notation, let’s take $m=p^2,c=2p\text{,}$ and $k=p^2{q}^2+1\text{,}$ where $p>0$ and $q>0\text{.}$ Then the complementary function of Eq. (1) is

We will take $p=5,q=3\text{,}$ and thus investigate the transient and steady periodic solutions corresponding to

with several illustrative possibilities for the external force F(t). For your personal investigations to carry out similarly, you might select integers p and q with $6\leqq p\leqq 9$ and $2\leqq q\leqq 5$.

Investigation 1: With periodic external force $F(t)=901\cos 3t\text{,}$ the Matlab commands

x = dsolve(′25*D2x+10*Dx+226*x=901*cos(3*t)′,
     ′x(0)=0, Dx(0)=0′);
x = simple(x);
syms t, xsp = cos(3*t) + 30*sin(3*t);
ezplot(x, [0  6*pi]),hold on
ezplot(xsp, [0  6*pi])

produce the plot shown in Fig. 5.6.13. We see the (transient plus steady periodic) solution

rapidly “building up” to the steady periodic oscillation $x_{\text{sp}}(t)=\cos 3t+30\sin 3t.$.

Investigation 2: With damped oscillatory external force

we have duplication with the complementary function in (2). The Maple commands

de2 := 25*diff(x(t),t,t)+10*diff(x(t),t)+226*x(t) =
       900*exp(-t/5)*cos(3*t);
dsolve({de2,x(0)=0,D(x)(0)=0}, x(t));
x := simplify(combine(rhs(%),trig));
C := 6*t*exp(-t/5);
plot({x,C,-C},t=0..8*Pi);

produce the plot shown in Fig. 5.6.14. We see the solution

oscillating up and down between the envelope curves $x=\pm 6t{e}^{-t/5}.$ (Note the factor of t that signals a resonance situation.)

Investigation 3: With damped oscillatory external force

we have a still more complicated resonance situation. The Mathematica commands

de3 = 25 x″[t] + 10 x′[t] + 226 x[t] ==
      2700 t Exp[-t/5] Cos[3t]
soln = DSolve[{de3, x[0] == 0, x′[0] == 0}, x[t], t]
x = First[x[t]  /.   soln]
amp = Exp[-t/5] Sqrt[(3t)^2 + (9t^2 - 1)^2]
Plot[{x, amp, -amp}, {t, 0, 10 Pi}]

produce the plot shown in Fig. 5.6.15. We see the solution

oscillating up and down between the envelope curves